Bounding the remainder I need to find the 3rd order Taylor polynomials and bound the remainder term at $(0,0)$.
The function is $$f(x,y)=\cos(x)\sin(y)$$
Here is what I did: first, I found the taylor expansions of sin and cos functions and multiply them, and then I get the following remainder term
$$R_3(x)-\frac{x^2}{2!}R_3(y)+yR_3(x)-\frac{y^3}{3!}R_3(x)+R_3(y)R_3(x)$$
I need to be using the inequality $R_{n,a}(x)\le \frac{M}{(n+1)!}|x-a|^{(n+1)}$
How do I proceed?
This is a pretty standard problem so I would be thankful if someone could give a methodological answer.
Thanks in advance!
 A: Lets refer to a set of notes for Multivariable Taylor and MV Taylor.
Lets do it the long way first.


*

*$(a, b) = (0, 0)$

*$f(x, y) = \cos x \sin y$

*$(x + y)^1 = x + y$, so we need partials $f_x, f_y$

*$f_{x}(x, y) = -\sin x \cos y \rightarrow f_x(0, 0) = 0$

*$f_{y}(x, y) = -\cos x \cos y \rightarrow f_y(0, 0) = 1$

*$(x + y)^2 = x^2 + 2 xy + y^2$, so we need partials $f_{xx}, f_{xy}, f_{yy}$

*$f_{xx}(x, y) = -\cos x \sin y \rightarrow f_{xx}(0, 0) = 0$

*$f_{xy}(x, y) = -\sin x \cos y \rightarrow f_{xy}(0, 0) = 0$

*$f_{yy}(x, y) = -\cos x \sin y \rightarrow f_{yy}(0, 0) = 0$

*Notice how all the second order terms were zero? This will be a pattern.

*$(x + y)^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3$, so we need partials $f_{xxx}, f_{xxy}, f_{xyy}, f_{yyy}$

*$f_{xxx}(x, y) = \sin x \sin y \rightarrow f_{xxx}(0, 0) = 0$

*$f_{xxy}(x, y) = -\cos x \cos y \rightarrow f_{xxy}(0, 0) = -1$

*$f_{xyy}(x, y) = \sin x \sin y \rightarrow f_{xyy}(0, 0) = 0$

*$f_{yyy}(x, y) = -\cos x \cos y \rightarrow f_{yyy}(0, 0) = -1$ 

*$(x + y)^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$, so we need partials $f_{xxxx}, f_{xxxy}, f_{xxyy}, f_{xyyy}, f_{yyyy}$

*However, we already detected a pattern that these will all be zero. This means that this error term is not enough and we need one more.

*$(x + y)^5 = x^5 + 5 x^4 y + 10x^3y^2+10x^2y^3+5 xy^4 + y^5$


We are now ready to calculate the linear approximation and the remainder term.
We have:
$$L(x, y) = L_1 (x, y) + \dfrac{1}{2!} L_2(x, y) + \dfrac{1}{3!} L_3(x, y)$$


*

*Using $(x + y)^1$

*$~~L_1(x, y) = f_x(a, b) (x - a) + f_y(a, b) (y - b) = 0 + y$

*Using $(x + y)^2$

*$~~L_2(x, y) = f_{xx}(a, b)(x-a)^2 + 2 f_{xy}(a, b)(x-a)(y-b) + f_{yy}(a,b)(y-b)^2 = 0$

*Using $(x + y)^3$

*$~~L_3(x, y) = f_{xxx}(a, b)(x-a)^3 + 3 f_{xxy}(a, b)(x-a)^2(y-b) +3 f_{xyy}(a,b)(x-a)(y-b)^2 + f_{yyy}(a, b)(y-b)^3 = 0 + 3(-1)x^2 y + 0 + (-1)y^3 = -3x^2y - y^3$


$$L(x, y) = y + \dfrac{1}{3!}(-3x^2 y - y^3) = y - \dfrac{x^2 y}{2} - \dfrac{y^3}{6}$$


*

*We already know $L_4 = 0$, so need one more term for the remainder, from $(x + y)^5$ we have:

*$R_5(x, y) = f_{xxxxx}(a, b)(x-a)^5 + 5f_{xxxxy}(a, b)(x-a)^4(y-b) + 10 f_{xxxyy}(a, b)(x-a)^3(y-b)^2 + 10 f_{xxyyy}(a, b)(x-a)^2(y-b)^3+5_{xyyyy}(a, b)(x-a)(y-b)^4+f_{yyyyy}(a, b)(y-b)^5$


As for bounding that error, we have that the max of the products for each of the partials of cosine and sine terms is $M = 1$, thus we can write:
$$|R_5(x, y)| \le \dfrac{M}{5!}(|x^5| + 5|x^4y| + 10|x^3 y^2| + 10 |x^2y^3|+ 5|x y^4| + |y^5|) = \dfrac{1}{120}(|x|+|y|)^5$$
Now, if they provided a region for $(x, y)$, we can easily calculate the max error.
There is a much easier approach to arrive at this solution.
The third order McLaurin series expansion for $\cos x = 1-\frac{x^2}{2}$ and for $\sin y = y-\frac{y^3}{6}$. This gives (see notes for dropped term):
$$L(x, y) = \cos x \sin y= \left(1-\frac{x^2}{2}\right) \left(y-\frac{y^3}{6}\right) = y -\frac{x^2 y}{2}-\frac{y^3}{6}$$
Compare that to the result above.
For the error term, we have the expansion for $(x + y)^5$. (When using this, be careful to know the max of the function because you need that value (see the sets of notes I linked above for examples) as I showed above for this particular example.
